1/14 Initials: Problem 4 - 25 points Two blocks of masses m, and my are x1 M attached to both ends of an ideal rope that is wrapped around a non-ideal pulley. The L_TR pulley is a solid cylinder of mass M and radius R. You may assume that there is no friction between block 1 and the table, the rope does not slip against the pulley, and the pulley M2 doesn't experience any friction while spinning around its axle. The acceleration due to gravity X2 has magnitude g. Note that the tension is not the same on both sides of the pulley. Take T1 on the left and T2 on the right of the pulley. a- Draw the complete free body diagram for each block and for the pulley. Use the coordinate systems and sign convention for rotation as indicated in the diagram. b- Write all the scalar equations resulting from Newton's second law for each block and for the pulley, as well as the relationship between the linear acceleration of each block and the angular acceleration of the pulley.Initials: C- Determine the magnitudes Ti and T2 of the two tension forces. d- Determine the magnitude of the acceleration of block 2.Initials: Physics 8A MT2 Equation Sheet r (t ) = x(t)ity(t)j Ar = 12- r1 = Axi+ Ayj Newton's Ist Law AT B = constant = ) F = o At VAV = = VAV-xi + VAV-yj Newton's 2nd Law D = lim = = Vxi + Vyj 4t-0 4t F = ma av aAV = At -= aAV-xi + aAV-y] Newton's 3rd Law a = lim = axi + ayj OFMe/You = -Fyou/Me v (t ) = vo + alt - to] Tension: T r(t) = ro+ volt- tol + alt - to] Ax = X2 - X1 Ax _X2 - X1 Normal Force VAV-x At t2 - t1 Ax dx Vx = li At-o At dt Avx Vx2 - Vx1 Gravity aAV-x At t2 - 1 Fg = mg AVX - dvx in (904) ax = lim - dt Static Friction Vx (t) = Vox + ax[t - to] 1 fs SUSN x(t) = xo + Vox[t - tol+ = ax[t - to]2 Kinetic Friction v2 = Vox + 2ax[x - xo] fK = HRN Ay = y2 - V1 Ay _y2 - y1 VAD - y Spring Force At t2 - t1 Ay dy Fsp = - kx (x measured from relaxed position) Vy = lim - At-0 At dt Avy Vy2 - Vy1 Uniform Circular Motion aAV-y At t2 - t1 Avy dvy a = ay = lim - At- 0 At dt v = rw Vy (t) = Voy + dy[t - to] T = - y(t) = yo + Voy[t - to] + 5 ay[t - to]2 vz = vay + 2ayly - yo]Initials: Work and Energy amounts B [ Fext = 0 = AP = 0 W = F . dr m1 - mz 2m2 V1f = 2 vjit- - Vzi my + mz mi + mz W = FL cosOF (constant angle and magnitude) 2m1 Vli +- m2 - m1, V2f = - Vzi Wg = -mg4h (height axis pointing upward) mit m2 mi + mz WSp = -(x3 - XA) (x measured from relaxed position) Rotational Kinematics 1 K = =mv2 0(t) = angular position 40 = 02 - 01 Wnet = > Wi = 4K 40 02 - 01 WAV = At t2 - t1 U + K =E de AE = WNC and AU = -Wc w = lim At-0 At at Ug = mgh (height axis pointing upward) AW W2 - W 1 a AV Usp = = kx2 (x measured from relaxed position) At t2 - t1 Aw dw a = lim - At-0 4t dt Universal Gravitation w (t ) = wot alt - tol FG = GMm 0 (t) = 00 + wolt - to] +5 alt- to]? GM g = - 7 = wg + 2a[0 -Oo] GMm s = ro (arc length) UG = = rw atan = ra GM 2 2 Vorb = arad = - =rw2 Rotational Dynamics Mys 2173/2 T =- VGM I = Et=1mir? (point masses) Ip = ICM + Md2 (parallel axis theorem) 2GM It| = rF sino notVesc = [Text = 1a Linear Momentum p = mi P = > Pi - Quadratic Equation [ Fest = at ax2 + bx + c = 0 has the solutions X = 2a -(-bvb2 - 4ac)Derivatives d(x12) Initials: = nx -1 dx Integrals dx = X2 - X1 Lengths, areas and volumes Circumference of circle: 2nR Area of disk: TR Surface area of sphere: 4TR- Lateral area of cylinder: 2ntRh Volume of cylinder: ntR h Volume of sphere: 4TR3/3 Trigonometry sin (90) = cos (09) = - cos (1809) = 1 sin (0) = cos (90) = sin (180) = 0 cos (60) = sin (30) = 1/2 sin (60) = cos (309) = V3/2 sin (450) = cos (450) = 12/2 cos (180- 0) = - cose sin (180- 0) = sine cos (90 + 0) = - cos (90 - 0) = - sine sin (90 + 0) = sin (90 - 0) = cose cos' 0 + sin- 0 = 1Initials: Table of rotational inertias M is the mass of the system Axis Axis Axis Hoop about central axis Annular cylinder Solid cylinder (or ring) about (or disk) about central axis central axis 1=MR2 (a) 1 = *M(R, 2 + R2?) ( b ) 1 = 4MR (C) Axis Axis Axis Solid cylinder Thin rod about Solid spherea (or disk) about axis through center bout any central diameter perpendicular to diameter length 2P 1= +MR + + ML2 (P) 1= LML? (e) 1 = 3MR (D) Axis Axis Axis Thin Hoop about any Slab about spherical shella diameter perpendicular bout any axis through 2R diameter center 1 = 1MR2 (8) 1 =MR (h ) 1 = 12 M(al + b? ) (1)